Saturday, August 23, 2014

9702 November 2011 Paper 41 42 Worked Solutions | A-Level Physics

  • 9702 November 2011 Paper 41 & 42 Worked Solutions | A-Level Physics

Paper 41 & 42


Question 1
Moon in circular orbit of radius r about a planet. Angular speed of moon in orbit = ω. Planet and moon considered to be point masses isolated in space.
Show that r3ω2 = constant:
The gravitational force provides the centripetal force.
GMm / r2 = mrω2      
r3ω2 = GM = constant since GM is a constant

Phobos and Deimos: moons in circular orbits about planet Mars.
Radius of orbit of Phobos: 9.39x106m
Radius of orbit of Deimos: 1.99x107m
Period of rotation of Phobos about Mars: 7.65hours
Mass of Mars:
For Moon Phobos,
ω = 2π / (7.65x3600) = 2.28x10-4rads-1
(9.39x106)3 x (2.28x10-4)2 = (6.67x10-11) x M
M = 6.46x1023 kg

Period of Deimos in orbit about Mars:
(9.39x106)3 x (2.28x10-4)2 = (1.99x107)3 x ω2
ω = 7.30x10-5 rads-1
T = 2π / ω = 2π / (7.30x10-5) = 8.6x104s = 23.6hours
Period of rotation of Mars about its axis = 24.6hours. Deimos: in an equatorial orbit, orbiting in same direction as spin of Mars about its axis.
Use answer from (i) to comment on orbit of Deimos:
Either Deimos is almost in a ‘geostationary’ orbit
Or Satellite would take a long time to cross sky

Question 2
1 assumption of kinetic theory of gases is that (gas) molecules behave as if they are hard, elastic identical spheres.
2 other assumptions of kinetic theory of gases:
The molecules/atoms/particles move in random (rapid) motion.
There are no intermolecular forces of attraction/ repulsion (between the molecules)
The volume of the molecules/atoms/particles is negligible compared to the volume of the container.
The time of collision (of the gas molecules) is negligible to the time between the collisions.

Using kinetic theory of gases, it can be shown that product of pressure p and volume V of an ideal gas is given by the expression             pV = (1/3) Nm<c2> where m is mass of a gas molecule.
Meaning of:
N: Number of (gas) molecules

<c2>: Mean square speed/velocity (of the gas molecules)

Use expression to deduce that mean kinetic energy <Ek> of gas molecule at temperature T is given by the equation            <Ek> = (3/2)kT            where k : constant
Either pV = NkT         or pV = nRT and links n and k
And <Ek> = ½ m<c2>
Any clear algebra leading to <Ek> = (3/2) kT

The internal energy of a substance is the sum of potential energy and kinetic energy of its molecules/atoms/particles that are randomly distributed.

Use <Ek> = (3/2) kT to explain that, for an ideal gas, a change in internal energy ΔU is given by    ΔU ΔT            where ΔT is change in temperature of gas:

No intermolecular forces are present in an ideal gas, so there is no potential energy. So, the (change in) internal energy is the (change in) kinetic energy and this is proportional to the (change in) temperature T.  

Question 3
Bar magnet suspended from free end of helical spring. One pole of magnet situated in coil wire. Coil connected in series with switch and resistor. Switch open. Magnet displaced vertically and then released. As magnet passes through its rest position, timer is started. Variation with time t of vertical displacement y of magnet from rest position shown. At t=4s, switch is closed.
Use Fig 3.2:
Evidence for magnet to be undergoing free oscillations during the period t = 0s to t = 4.0s:
The amplitude remains constant.

Reason whether damping after t = 4s is light, critical or heavy:
The amplitude decreases gradually. So, damping is light.

Determine natural frequency of vibration of magnet on spring:
Period of oscillations, T = 0.80s
So, natural frequency of vibration = 1/T = 1/0.80 = 1.25Hz

Faraday’s law of electromagnetic induction states that the (induced) e.m.f is proportional to the rate of change/cutting of (magnetic) flux (linkage).

Why, after time t = 4.0s, amplitude of vibration of magnet is seen to decrease:
A current is induced in the coil as the magnet moves in the coil. The current in the resistor gives rise to a heating effect. This thermal energy is derived from the energy of oscillation of the magnet.

Question 4
{Detailed explanations for this question is available as Solution 609 at Physics 9702 Doubts | Help Page 120 -}

Question 5
+vely charged particles are travelling in vacuum through 3 narrow slits S1, S2 and S3. Each particle: speed v and charge q. Uniform magnetic flux density B and uniform electric field of field strength E in region between slits S2 and S3.
Expression for force F acting on a charged particle due to
Magnetic field:
Bqv(sinθ)        or Bqv(cosθ)

Electric field:

Electric field acts downwards in plane of paper. Direction of magnetic field so that the +ve charged particles may pass undeviated through region between slits S2 and S3:
The direction of the force, FB provided by the magnetic field must be opposite to the direction of the force due to the electric field. So, the magnetic field must be into the plane of the paper.

Question 6
Variation with time t of output V of alternating voltage supply of frequency 50Hz.
Use fig 6.1 to state:
Time t1:
Period, T = 1/frequency = 1/ 50 = 0.02s
So, t1 = 0.03s
Peak value Vo of voltage:
Peak value of voltage = 17.0V

Root-mean-square voltage Vrms:
Vrms = 17.0 / √2 = 12.0V

Mean voltage <V>:
<V> = 0

Alternating supply connected in series with resistor of resistance 2.4Ω. Mean power dissipated in resistor:
Power dissipated = V2/R = 122/2.4 = 60W

Question 7
{Detailed explanations for this question is available as Solution 599 at Physics 9702 Doubts | Help Page 118 -}

Question 8
Isotope phosphorus-33 undergoes beta-decay to form sulfur-33, which is stable. Half-life of phosphorus-33 = 24.8days.
Radioactive half-life is defined as the time for the initial number of nuclei/activity to reduce to one half of its initial value.

Show that decay constant of phosphorus-33 = 3.23x10-7s-1:
Decay constant, λ = ln2 / (24.8x24x3600) = 3.23x10-7s-1

Pure sample of phosphorus-33 has initial activity of 3.7x106Bq.
Initial number of phosphorus-33 nuclei in sample:
A = λN
3.76x106 = 3.23x10-7 x N
N = 1.15x1013

Number of phosphorus-33 nuclei remaining in sample after 30days:
N = Noexp(-λt) = 1.15x1013 x exp(-{[ln2] x 30}/24.8)
N = 4.97x1012

After 30 days, sample in (b) will contain phosphorus-33 and sulfur-33 nuclei. Ratio of number of phosphorus-33 nuclei after 30 days to number of sulfur-33 nuclei after 30 days:
Ratio = (4.97x1012) / ([1.15x1013] – [4.97x1012]) = 0.76


Question 9
2 effects of negative feedback on gain of amplifier incorporating an operational amplifier (op-amp):
Any 2 sensible suggestions:
Reduced gain
Greater bandwidth or less distortion
Increased stability

Circuit diagram of a non-inverting amplifier using ideal op-amp:
Complete diagram. Label input and output:
V- is connected to the midpoint between resistors
VOUT clear and input to V+ clear

Resistance R so that non-inverting amplifier has voltage gain of 15:
Voltage gain = 1 + RF/R
15 = 1 + (12000/R)
R = 860Ω

Draw graph to show variation with input potential VIN of output potential VOUT. Consider input potentials in the range 0 to +1.0V:
A straight line from point (0, 0) to (0.6, 9.0)
A straight line from point (0.6, 9.0) to (1.0, 9.0)

Output of amplifier circuit may be connected to a relay.
1 purpose of a relay:
A relay can be used to switch a large current/voltage since the output current of an op-amp is a few mA/very small.

A relay can be used as a remote switch for inhospitable region/avoid using long heavy cables.

Question 10
Cable television uses optic fibres for transmission of signals.
4 advantages of optic fibres over coaxial cables for transmission of data:
Any 4 sensible suggestions:
Large bandwidth/carries more information
Low attenuation of the signal
Low cost
Smaller diameter, easier handling, easier storage, less weight
High security/no crosstalk
Low noise/no EM interference

EM radiation of wavelength 1310nm frequently used for optic fibre communication, rather than visible light.
Region of EM spectrum in which radiation of wavelength 1310nm is found:
Infra-red region

Why this radiation is used, rather than visible light:
There is a lower attenuation for this radiation than for visible light.

An optic fibre has attenuation per unit length of 0.2dBkm-1. Signal transmitted along optic fibre of length 30km to receiver. Noise power at receiver = 9.3μW. Minimum acceptable signal-to-noise ratio at receiver = 26dB.
Minimum signal power at receiver:
Gain / dB = 10 log(P2/P1)
26 = 10 log (P2/(9.3x10-6))
So, P2 = 3.7x10-3W

Minimum input signal power to optic fibre:
Power loss along the fibre = 30 x 0.2 = 6.0dB
Either 6 = 10 log(P/(3.7x10-3))
Or 32 = 10 log(P/(9.3x10-6))
Minimum input power signal, P = 1.5x10-2W

Question 11
Simplified block diagram of mobile phone handset shown.
Name and function of:
Block A:
Switch: so that one aerial can be used for transmission and reception

Block B:
Tuning circuit: to select (one) carrier frequency (and reject others)

Block C
ADC/analogue-to-digital converter: converts the microphone output to a digital output

Block D:
(a.f.) amplifier: to increase the (power of) signal to drive the loudspeaker

2 reasons why communication between mobile phone handset and base station is conducted using UHF:
Any 2 sensible suggestions:
Short aerial in handsets so easy to handle
Short range so less interference between the base stations
A larger waveband so more carrier frequencies


  1. in question 7 c, why are only two dark lines seen?

    1. The details for question 7 has been updated. Check again

  2. Can you explain Q4? Why is (b)(ii) for maximum, the distance can't be 1.4 cm?

    1. Some more details have been added


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